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  "body": "\n## Statement\n\nLet $\\varphi$ be a [[mollifier|CQMK4YPE1J]] on $\\R^n$, let $\\varphi_\\varepsilon(x) = \\varepsilon^{-n} \\varphi(x/\\varepsilon)$ for $\\varepsilon \u003e 0$, and let $f$ be a locally integrable function on $\\R^n$.\nPut $f_\\varepsilon = \\varphi_\\varepsilon * f$, the [[convolution|2Z29JN3VES]] $f_\\varepsilon(x) = \\int_{\\R^n} \\varphi_\\varepsilon(x - y) f(y) \\, dy$.\nThen:\n\n1. $f_\\varepsilon$ is a [[smooth function|0V33BNKZ9S]], and $\\partial^\\alpha f_\\varepsilon = (\\partial^\\alpha \\varphi_\\varepsilon) * f$ for every multi-index $\\alpha$.\n2. If $f$ vanishes almost everywhere outside a closed set $K$, then $f_\\varepsilon$ vanishes outside $\\set{x : \\operatorname{dist}(x, K) \\le \\varepsilon}$.\n3. If $f$ is [[continuous|0X3WGX95FE]], then $f_\\varepsilon \\to f$ uniformly on every compact set as $\\varepsilon \\to 0$.\n4. If $f \\in L^p(\\R^n)$ with $1 \\le p \u003c \\infty$, then $\\norm{f_\\varepsilon}_{L^p} \\le \\norm{f}_{L^p}$ and $\\norm{f_\\varepsilon - f}_{L^p} \\to 0$ as $\\varepsilon \\to 0$.\n\n## Remarks\n\nPart 1 follows by differentiating under the integral sign, which is allowed because $\\varphi_\\varepsilon$ has bounded derivatives and compact support.\nPart 2 holds because $\\varphi_\\varepsilon(x - y) = 0$ when $\\abs{x - y} \u003e \\varepsilon$.\nFor parts 3 and 4, write $f_\\varepsilon(x) - f(x) = \\int \\varphi_\\varepsilon(y) \\big( f(x - y) - f(x) \\big) dy$, an average of differences over $\\abs{y} \\le \\varepsilon$.\nFor continuous $f$, uniform continuity on compact sets makes it small.\nFor $f \\in L^p$, Minkowski's integral inequality bounds its norm by $\\sup_{\\abs{y} \\le \\varepsilon} \\norm{f(\\cdot - y) - f}_{L^p}$.\nThis tends to $0$ by the continuity of translation in $L^p$.\nThe bound $\\norm{f_\\varepsilon}_{L^p} \\le \\norm{f}_{L^p}$ is [[Young's convolution inequality|JXDKJCBXJ8]] with $\\norm{\\varphi_\\varepsilon}_{L^1} = 1$.\n\nConsequence: for $1 \\le p \u003c \\infty$ the [[bump functions|HS65DW0HXH]] are dense in $L^p(\\R^n)$.\nIndeed, first cut $f$ off to a large ball, then mollify, and use parts 1, 2 and 4.\n\nPart 4 fails for $p = \\infty$.\nIf $f$ is the indicator function of $[0, \\infty)$ on $\\R$, then each $f_\\varepsilon$ is continuous, so $\\norm{f_\\varepsilon - f}_{L^\\infty} \\ge 1/2$ for all $\\varepsilon$.\n\n## Explanations\n\n### L1 human\n\nBlurring a function with a narrow smooth blob gives a smooth function.\nIf the blob is narrow enough, the blurred function is as close as one wants to the original.\nThe closeness is uniform for continuous functions and in the $L^p$ sense for $L^p$ functions.\nSo rough functions can be replaced by smooth ones, at a small cost.\n\n### L2 human\n\nThis theorem is the workhorse behind density arguments in analysis.\nTo prove an identity or an inequality for all $f \\in L^p$ or all $f$ in a [[Sobolev space|M4CEZ5K9HP]], one proves it for smooth functions and passes to the limit.\nMollification commutes with weak derivatives, $\\partial^\\alpha (\\varphi_\\varepsilon * f) = \\varphi_\\varepsilon * \\partial^\\alpha f$, so the same argument shows that smooth functions are dense in $W^{k,p}(\\R^n)$.\nAt a Lebesgue point of $f$, $f_\\varepsilon(x) \\to f(x)$, so the convergence also holds almost everywhere.\n",
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